Difference between revisions of "2022 AMC 10B Problems/Problem 2"
Ghfhgvghj10 (talk | contribs) (→Solution 2) |
|||
(17 intermediate revisions by 9 users not shown) | |||
Line 25: | Line 25: | ||
\textbf{(E) }25</math> | \textbf{(E) }25</math> | ||
− | ==Solution== | + | ==Solution 1== |
<asy> | <asy> | ||
Line 43: | Line 43: | ||
</asy> | </asy> | ||
− | + | <cmath>\textbf{Figure redrawn to scale.}</cmath> | |
<math>AD = AP + PD = 3 + 2 = 5</math>. | <math>AD = AP + PD = 3 + 2 = 5</math>. | ||
Line 49: | Line 49: | ||
<math>ABCD</math> is a rhombus, so <math>AB = AD = 5</math>. | <math>ABCD</math> is a rhombus, so <math>AB = AD = 5</math>. | ||
− | <math>\bigtriangleup APB</math> is a 3-4-5 right triangle, | + | <math>\bigtriangleup APB</math> is a <math>3-4-5</math> right triangle, hence <math>BP = 4</math>. |
− | The area of the rhombus <math> | + | The area of the rhombus is base times height: <math>bh = (AD)(BP) = 5 \cdot 4 = \boxed{\textbf{(D) }20}</math>. |
~richiedelgado | ~richiedelgado | ||
− | ==Solution 2== | + | ==Solution 2 (The Area Of A Triangle)== |
<asy> | <asy> | ||
pair A = (0,0); | pair A = (0,0); | ||
Line 73: | Line 73: | ||
</asy> | </asy> | ||
− | The diagram is | + | The diagram is from as Solution 1, but a line is constructed at <math>BD</math>. |
− | When it comes to the sides of a rhombus, their opposite sides are congruent and parallel. | + | When it comes to the sides of a rhombus, their opposite sides are congruent and parallel. This means that <math>\angle ABD \cong \angle BDC</math>, by the Alternate Interior Angles Theorem. |
− | By | + | By SAS Congruence, we get <math>\triangle ABD \cong \triangle BDC</math>. |
− | + | Since <math>AP=3</math> and <math>AB=5</math>, we know that <math>BP=4</math> because <math>\triangle APB</math> is a 3-4-5 right triangle, as stated in Solution 1. | |
− | + | The area of <math>\triangle ABD</math> would be <math>10</math>, since the area of the triangle is <math>\frac{bh}{2}</math>. | |
− | Since we know that <math>\triangle ABD | + | Since we know that <math>\triangle ABD \cong \triangle BDC</math> and that <math>ABCD=\triangle ABD + \triangle BDC</math>, so we can double the area of <math>\triangle ADB</math> to get <math>10 \cdot 2 = \boxed{\textbf{(D) }20}</math>. |
− | ~ghfhgvghj10 | + | ~ghfhgvghj10, minor edits by MinecraftPlayer404 |
+ | |||
+ | ==Video Solution 1== | ||
+ | https://youtu.be/Io_GhJ6Zr_U | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution(1-16)== | ||
+ | https://youtu.be/SCwQ9jUfr0g | ||
+ | |||
+ | ~~Hayabusa1 | ||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/_KNR0JV5rdI?t=97 | ||
== See Also == | == See Also == |
Latest revision as of 17:14, 29 June 2023
- The following problem is from both the 2022 AMC 10B #2 and 2022 AMC 12B #2, so both problems redirect to this page.
Contents
Problem
In rhombus , point
lies on segment
so that
,
, and
. What is the area of
? (Note: The figure is not drawn to scale.)
Solution 1
.
is a rhombus, so
.
is a
right triangle, hence
.
The area of the rhombus is base times height: .
~richiedelgado
Solution 2 (The Area Of A Triangle)
The diagram is from as Solution 1, but a line is constructed at .
When it comes to the sides of a rhombus, their opposite sides are congruent and parallel. This means that , by the Alternate Interior Angles Theorem.
By SAS Congruence, we get .
Since and
, we know that
because
is a 3-4-5 right triangle, as stated in Solution 1.
The area of would be
, since the area of the triangle is
.
Since we know that and that
, so we can double the area of
to get
.
~ghfhgvghj10, minor edits by MinecraftPlayer404
Video Solution 1
~Education, the Study of Everything
Video Solution(1-16)
~~Hayabusa1
Video Solution by Interstigation
https://youtu.be/_KNR0JV5rdI?t=97
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 1 |
Followed by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.